Timeline for How to solve for $a$ in $\sum_{j=i}^n (a -j) \binom{n}{j} y^j (1-y)^{n-j} = 0$

11 events

when toggle format	what		by	license	comment
Jun 29, 2022 at 18:43	vote	accept	E. Turok
Jun 26, 2022 at 21:53	comment	added	Max Alekseyev		@E.Turok: Use the property $j\binom{n}{j}=n\binom{n-1}{j-1}$.
Jun 26, 2022 at 21:32	comment	added	E. Turok		@MaxAlekseyev how did you remove the $j$ in the numerator sum? Note that $$ B_{i, n} = \sum_{j=i}^n \binom{n}{j} y^j (1-y)^{n-j} \neq \sum_{j=i}^n j \binom{n}{j} y^j (1-y)^{n-j} $$
Jun 26, 2022 at 21:31	history	edited	E. Turok	CC BY-SA 4.0	edited body
Jun 26, 2022 at 21:27	comment	added	Max Alekseyev		Possible simplification for $i>0$: $$a = \frac{nyB_{i-1,n-1}(y)}{B_{i,n}(y)}.$$
Jun 26, 2022 at 21:27	answer	added	Carlo Beenakker		timeline score: 1
Jun 26, 2022 at 21:06	history	edited	E. Turok	CC BY-SA 4.0	added 333 characters in body
Jun 25, 2022 at 3:21	comment	added	Brendan McKay		Do you want a theoretical solution, or one that is practical for small $n$? $a$ is the expectation of a binomial random variable with parameters $n,y/(1-y)$, conditioned on being at least $i$. I don't think there is an exact closed form except in terms of special functions like hypergeometric functions.
Jun 24, 2022 at 18:39	comment	added	Qiaochu Yuan		I don't understand the question. Isn't this a linear equation in $a$?
S Jun 24, 2022 at 17:02	review	First questions
Jun 24, 2022 at 18:46
S Jun 24, 2022 at 17:02	history	asked	E. Turok	CC BY-SA 4.0